http://acm.hdu.edu.cn/showproblem.php?pid=1050

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Problem Description

The famous ACM (Advanced Computer Maker) Company has rented a floor of a building whose shape is in the following figure. 

The floor has 200 rooms each on the north side and south side along the corridor. Recently the Company made a plan to reform its system. The reform includes moving a lot of tables between rooms. Because the corridor is narrow and all the tables are big, only one table can pass through the corridor. Some plan is needed to make the moving efficient. The manager figured out the following plan: Moving a table from a room to another room can be done within 10 minutes. When moving a table from room i to room j, the part of the corridor between the front of room i and the front of room j is used. So, during each 10 minutes, several moving between two rooms not sharing the same part of the corridor will be done simultaneously. To make it clear the manager illustrated the possible cases and impossible cases of simultaneous moving. 

For each room, at most one table will be either moved in or moved out. Now, the manager seeks out a method to minimize the time to move all the tables. Your job is to write a program to solve the manager’s problem.

Input

The input consists of T test cases. The number of test cases ) (T is given in the first line of the input. Each test case begins with a line containing an integer N , 1<=N<=200 , that represents the number of tables to move. Each of the following N lines contains two positive integers s and t, representing that a table is to move from room number s to room number t (each room number appears at most once in the N lines). From the N+3-rd line, the remaining test cases are listed in the same manner as above.

Output

The output should contain the minimum time in minutes to complete the moving, one per line.

Sample Input

3
4
10 20
30 40
50 60
70 80
2
1 3
2 200
3
10 100
20 80
30 50

Sample Output

10
20
30

Problem solving report:

Description: 要在一個走廊搬桌子，從第s號房間到第t號房間，可以多組一起搬，因為走廊的地方有限，不能同時搬重合的，例如10-20,30-40可以一起搬，但是10-20,15-30不能同時搬，求總計用多少時間能搬完桌子(最短時間)。
Problem solving:

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
int main()
{
    int tt, t, n, s, a[220], max;
    scanf("%d", &tt);
    while (tt--)
    {
        scanf("%d", &n);
        memset(a, 0, sizeof(a));
        max = 0;
        for (int i = 0; i < n; i++)
        {
            scanf("%d%d", &s, &t);
            if (s > t)
                swap(s, t);
            for (int j = (s + 1) / 2; j <= (t + 1) / 2; j++)//(s+1)/2表示第幾個走廊空間，例如，房間1門前的是1，房間3門前的是2...
            {
                a[j]++;
                if (max < a[j])
                    max = a[j];
            }
        }
        printf("%d\n", max * 10);
    }
    return 0;	
}

②方法二：可以用貪心中的不相交區間來做，我們可以把每次不相交的標記一下，統計一下搬幾次就行了。

#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
struct edge {
    int left, right;
}e[205];
bool cmp(edge a, edge b) {
    return a.left < b.left;
}
int main()
{
    int tt, n, s, t, ans, right, vis[205];
    scanf("%d", &tt);
    while (tt--)
    {
        scanf("%d", &n);
        for (int i = 0; i < n; i++)
        {
            scanf("%d%d", &s, &t);
            if (s > t)
                swap(s, t);
            e[i].left = (s + 1) / 2;//每個房間門前走廊的編號，房間1和房間2是一樣的
            e[i].right = (t + 1) / 2;
        }
        ans = 0;
        sort(e, e + n, cmp);
        memset(vis, 0, sizeof(vis));
        for (int i = 0; i < n; i++)
        {
            if (vis[i])//已經搬過了
                continue;
            ans++;//又開始一次
            vis[i] = 1;//標記一下
            right = e[i].right;
            for (int j = i + 1; j < n; j++)
            {
                if (!vis[j] && right < e[j].left)//是否相交
                {
                    vis[j] = 1;
                    right = e[j].right;
                }
            }
        }
        printf("%d\n", ans * 10);
    }
    return 0;
}